What will be the output of the following code?class Node: def __init__(self, data=None): self.data = data self.next = Noneclass LinkedList: def __init__(self): self.head = None def print_list(self): current_node = self.head while current_node: print(current_node.data) current_node = current_node.next def insert_at_end(self, new_data): new_node = Node(new_data) if self.head is None: self.head = new_node return last_node = self.head while last_node.next: last_node = last_node.next last_node.next = new_nodellist = LinkedList()llist.insert_at_end(1)llist.insert_at_end(2)llist.insert_at_end(3)llist.print_list()Options321132123213
Question
What will be the output of the following code?class Node: def init(self, data=None): self.data = data self.next = Noneclass LinkedList: def init(self): self.head = None def print_list(self): current_node = self.head while current_node: print(current_node.data) current_node = current_node.next def insert_at_end(self, new_data): new_node = Node(new_data) if self.head is None: self.head = new_node return last_node = self.head while last_node.next: last_node = last_node.next last_node.next = new_nodellist = LinkedList()llist.insert_at_end(1)llist.insert_at_end(2)llist.insert_at_end(3)llist.print_list()Options321132123213
Solution
The output of the code will be:
1 2 3
This is because the insert_at_end method of the LinkedList class inserts new nodes at the end of the linked list. So, when we insert 1, 2, and 3 in that order, they will be printed in the same order by the print_list method.
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